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Ximeng
China57 Posts
You are going for a large straight (5 dice in sequence) you have 3 rolls. You may keep any dice between rolls. Your first roll comes up as a 5 3 3 2 1. With 2 rolls left is it better to: a) roll the extra 3 and try to change it to a 4 to complete a sequence of 5 to 1 b) roll the extra 3 and the 1 to try to complete a sequence of 6 to 2 or 5 to 1 What is the probability of success in each case? The answer is actually quite surprising.  | ||
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Jumbled
1543 Posts
Edit: On second thoughts, there's some double counting in that 35% figure. It's actually going to be less than that. Yup, without the double-counting it comes to 27.777...% | ||
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ghrur
United States3786 Posts
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Iranon
United States983 Posts
Chance of this with 2 shots it: (1/6)+(5/6)(1/6) = 30.56% Chance of this with 2 shots at it: (4/36)+(32/36)(4/36) = 13.58% Edit: blah, my second case is wrong, you should also factor in the chance that you get either a 1/4/6 on the first try and the other number you need on the second try. Going to sleep instead of writing it out. Edit 2: Can't sleep, might as well fix this. We're rolling up to 2 dice up to 2 times, trying for either a 4 and a 6 or a 4 and a 1: On your first roll, you should keep anything with one of those three numbers, or 11, 12, 13, 14, 15, 16, 21, 24, 26, 31, 34, 36, 41, 42, 43, 44, 45, 46, 51, 54, 56, 61, 62, 63, 64, 65, 66 In four of those twenty-seven possibilities (14, 41, 46, 64), you're done. In 7 of the others (24, 34, 42, 43, 44, 45, 54) you have a 4 and a useless number -- on your next roll you need a 1 or a 6 (1/3). In the other sixteen, you have a 1 or a 6 and a useless number -- on your next roll you need a 4 (1/6). If you don't roll any of those twenty-seven, then on your second roll you need to hit 14, 41, 46, or 64 (4/36). So overall, the probability for the second option is (4/36)+(7/36)(1/3)+(16/36)(1/6) + (9/36)(4/36) = 27.78% This is less than, but surprisingly close to, the probability of doing the first option. | ||
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n.DieJokes
United States3443 Posts
For one die, its 1/6 +1/6. So 1/3 chance. The other one is trickier: Both win: 2*(1/3)(1/3) * (36/36) because there are two ways One win one lose: 2*(2/3)(1/3)*(1/6) Both lose: (2/3)(2/3)*(2/36)*2 Add em all up! .37 and some change for option two, Thus its better Edit: For more fun probability problems theres a super cheap dover book called like "50 interesting problems in probability with solutions" | ||
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Jumbled
1543 Posts
On July 27 2012 13:19 n.DieJokes wrote: + Show Spoiler [solution] + For one die, its 1/6 +1/6. So 1/3 chance. The other one is trickier: Both win: 2*(1/3)(1/3) * (36/36) because there are two ways One win one lose: 2*(2/3)(1/3)*(1/6) Both lose: (2/3)(2/3)*(2/36)*2 Add em all up! .37 and some change for option two, Thus its better Edit: For more fun probability problems theres a super cheap dover book called like "50 interesting problems in probability with solutions" You may want to redo that, your numbers are incorrect. | ||
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n.DieJokes
United States3443 Posts
On July 27 2012 13:57 Jumbled wrote: You may want to redo that, your numbers are incorrect. How do you figure? | ||
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RAGEMOAR The Pope
United States216 Posts
For #1 1/6 chance that you win THEN 5/6 chance that 1/6 win 11/36 chance. | ||
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n.DieJokes
United States3443 Posts
On July 27 2012 14:14 RAGEMOAR The Pope wrote: For #1 1/6 chance that you win THEN 5/6 chance that 1/6 win 11/36 chance. Ah true, whoopsies | ||
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RAGEMOAR The Pope
United States216 Posts
Same applies to your second case, except much, much more wrong. | ||
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n.DieJokes
United States3443 Posts
On July 27 2012 14:17 RAGEMOAR The Pope wrote: Same applies to your second case, except much, much more wrong. Oh well | ||
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Gnaix
United States438 Posts
On your first roll you roll either (6, 4), (4, 6), (4, 1), (1, 4), in which case you don't need to roll the second time. Probability: 4/36 On your first roll, one of your die rolls a 6 and the other rolls not a 4, in which case you keep the 6 die and you roll the other die a second time for a 4. There are 4*2+1 ways, since 6 paired with any other number can happen 2 ways but only 1 way with itself. Probability: (4*2+1)/36*1/6 On your first roll, one of your die rolls a 1 and the other rolls not a 4, in which case you keep the 1 die and you roll the other die a second time for a 4. There are 4*2+1 ways. Probability: (4*2+1)/36*1/6 On your first roll, one of your die rolls a 4 and other other rolls not a 6 or 2, in which case you keep the 4 die and roll the other die a second time for a 2 or 6. There are 3*2+1 ways. Probability: (3*2+1)/36*1/3 On your first roll, you don't get any of the above, therefore you need to get (6,4) (4,6) (4,1) (1,4) on your second roll. There are 3*3 ways. Probability: (3*3)/36*4/36 Total probability is the sum of the probabilities above which is around 0.29 | ||
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Iranon
United States983 Posts
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endy
Switzerland8970 Posts
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Helge
Ukraine4 Posts
2) A bit more complicated calculation. On your first try, out of 6*6=36 possible rolls, you can either make: - perfect rolls = instant win. You need to roll either (1,4) or (6,4) or (4,1) or (4,6). The chance is 4/36 = 1/9. - good rolls = have 4 but not a perfect one. 7 possible rolls: (4,2),(4,3),(4,5),(4,4),(2,4),(3,4),(5,4). The chance is 7/36 - awful rolls = no 1,4 or 6 on any dices. There are 3*3=9 such rolls. The chance is 9/36 = 1/4. - questionable rolls - have 1 or 6 but no 4. All rolls that remains, so there are 36-4-7-9=16 of them. The chance is 16/36=4/9. Your second try depends on your first try: - perfect roll - you already won, the overall chance is 1/9, or approx. 11,1% - good roll. Keeping 4 let's you win in 2 out of 6 rolls (1 or 6), which is much better that rolling 2 dices again. So the overall chance is (7/36)*(2/6) = approx. 6,5% - awfull roll. Need to roll perfect second time, so the chance is = (1/4)*(1/9)= approx. 2,8% - questionable roll, you can either keep 1 dice (1 or 6) or reroll both of them. In the first case your chances are 1/6 (roll 4 on 1 dice), or 1/9 (a perfect roll as described above). So naturally you'll want to roll 1 dice. So yout overall chance is (4/9)*(1/6)= approx. 7,4% Total chance is 11,1%+6,5%+2,8%+7,4% = 27,8% (rounding errors are about 0.1%). So the first strategy is better. | ||
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KillerDucky
United States498 Posts
http://www-set.win.tue.nl/~wstomv/misc/yahtzee/osyp.php For this situation, you can put a 0 in for everything except the Large Straight to force it to consider that option. Give it 53321 as roll #1, and it says: #1 Keep 53_21, roll 3 = EV 12.22 #2 Keep 53_2_, roll 3,1 = EV 11.11 Since you're going to succeed and get 40 points, or fail and get 0 points, you can calculate probability for success: #1 12.11/40 = 30.275% #2 11.11/40 = 27.775% | ||
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32
United States163 Posts
http://pastebin.com/mXCjB1Uq I might be doing it wrong but I don't think so. The entries with a + sign are added to the total probability, and I do keep track of the probability of the roll when I add it to the total. I ended up with 25.3086419753%, I think python has negligible rounding errors though. Only rolling the second 3 again http://pastebin.com/KxTA83f9 | ||
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